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So, what you've got here is base current from the lower transistor powering the LED. Even if the voltage at its collector is 0, the source you have feeding the 10k resistor is delivering current to the LED through the base-emitter junction of the transistor. This means that the LED will light, even with no upper transistor. That's why the LED is so dim too.

When you pull both bases up, the upper transistor supplies voltage to the collector of the lower one, which allows it to source more current to the LED, making it brighter. But if the lower transistor's base is pulled low, it cuts off all current to the LED.

If you want to make an AND gate with a pair of BJTs, it's not feasible to do it like this, at least not with the values used.

Let's assume the voltage that you are apply to the gate inputs is exactly 5V for argument's sake. Let's take Vbe (the diode drop between the base and emitter of the transistors) as 0.6V. When you pull the bottom gate input high, the bottom transistor (saturated or not, but hopefully saturated) will have about 4.4 volts on it's emitter.

Now the problem is that Vce(sat) of the bottom transistor is not zero. It is perhaps 0.1V. This means that the voltage on the collector of the bottom transistor is 4.5V (or that is what it will be when the top transistor is ever turned on).

So, now we apply a 5V signal to the top gate input and now we have 5V on the top transistor base. So, now the base voltage of the top transistor is only 0.5V above the voltage on the emitter.

Since the voltage we are apply across the base-emitter junction of the top transistor is less than a diode drop, we can never turn on the top transistor under these conditions.

The problem is that the circuit on the school's webpage is a conceptual representation but is not a practical circuit.

I think that the circuit as-is can work as an AND gate if the active "high" input on the bottom transistor is fudged so as to only be 4.5V and the "high" input of the top transistor is implemented as the full 5V. This will provide sufficient bias clearance for the top transistor to turn on properly Try adding a 100K resistor from the base of the bottom transistor to ground. That might be enough. The side effect is that the "high" output of the entire circuit will only be about 3.8 or 3.9V.

I think that the reason you see some LED illumination with just the bottom input "high" is because the b-e junction of the bottom transistor is acting like just a diode and you are seeing leakage base current through the LED.